A child bounces in a harness suspended from a door frame by three elastic bands. If each elastic band stretches 0.300 m while supporting a 7.15-kg child at rest, what is the force constant for each elastic band? Assume that each spring supports 1/3 of the child's weight.

Hooke's law of a spring or an elastic band gives the relation between elastic force (Fe) and stretching (x), the magnitude of that force is:

[tex] F_{e}= kx [/tex] (1)

With k, the elastic force constant. The three elastic bands support the child’s weight (W) and maintain him at rest, so by Newton’s second law for one of the elastic bands:

[tex] \sum F=0 [/tex] (2)

[tex]\frac{W}{3}-F_{e}=0\Rightarrow F_{e}=\frac{W}{3} [/tex] (3)

Using (1) on (3):

[tex] kx=\frac{W}{3}\Rightarrow k=\frac{mg}{3x} [/tex](4)

[tex] k=\frac{7.15\,kg*9.81\,\frac{m}{s^{2}}}{3*0.300\,m}\simeq\mathbf{77.93\frac{N}{m}} [/tex]

asuwafohjames asuwafohjames · Answer 2 · 2022-04-01T12:14:19+02:00

The force constant of each elastic band is 77.87 N/m.

To find the force constant for each elastic band, we use Hooke's law

Hooke's law

Hooke's law states that the force applied to an elastic material is directly proportional to its extension provided the elastic limit is not exceeded

W = ke............ Equation 1

Where:

W = Weight of the child on one elastic band
k = Force constant
e = extension

Make k the subject of the equation

k = W/e.............. Equation 2

From the question,

Given:

W = (7.15×9.8)/3 = 23.36 N
e = 0.3 m

Substitute these values into equation 2

k = 23.36/0.3
k = 77.87 N/m

Hence, The force constant of each elastic band is 77.87 N/m.
Learn more about hook's law here: https://brainly.com/question/12253978